PQRS is a square loop of a conducting wire. A current I enters the loop at Pand leaves the loop at S. The magnitude of magnetic field at the centre o of the square is

Options

A. `(mu_(0))/(4pi) (2sqrt2I)/(a)`
B. `(mu_(0))/(4pi) (4sqrt2I)/(a)`
C. `(mu_(0))/(4pi) (4sqrt2I)/(a)`
D. zero

Answer

D

Explanation

The resistence of ann PQRS is 3 times the resistence of arm Ps.If resistance of arm PS=R then resistance of arm PQRS=3R.Potential differnce across PQRS and PS will be equal as they are connected in parallel
`:.I_(1)R=I_(2)xx3R`
or " " `I_(1)=3I_(2)`
Magnetic field at O due to curent through arm PS is,
`B_(1)=(mu_(0))/(4pi)(I_(1))/(a"/"2)[sin45^(@)+sin45^(@)]`
acting perpendicular to the loop upwards
`=(mu_(0))/(4pi)(3I_(2))/((a"/"2))[sin45^(@)+sin45^(@)]`
Magnetic field at O due to current throough arm PQRS is
`B_(2)=(mu_(0))/(4pi)(I_(2))/((a"/"2))[sin45^(@)+sin45^(@)]xx3`
actiing perpendicular to the loop downward
`:.` Net magnetic field at O is `B=B_(1)-B_(2)=0`